Month: September 2024

JAC Board Class 9th Science Solutions Chapter 11 Work and Energy

JAC Class 9th Science Work and Energy InText Questions and Answers

Page 148

Question 1.
A force of 7N acts on an object. The displacement is, say 8m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer:
When a force F acts on an object to displace it through a distance s in its direction, the work done, W on the body by the force is given by:
Work done = Force × Displacement
W = F × s
Where, F = 7N, and s = 8m Therefore, work done,
W = 7 × 8 = 56 Nm = 56J

Page 149

Question 1.
When do we say that work is done?
Answer:
Work is said to be done whenever the following two conditions are satisfied:

1. A force acts on the body.
2. There is a displacement in the body caused by the applied force along the direction of the force.

Question 2.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
When a force ‘F’ displaces a body through a distance ‘s’ in the direction of the applied force, the work done, ‘W’ on the body is given by the expression:
Work done = Force × Displacement W = F × s

Question 3.
Define 1J of work.
Answer:
1J is the amount of work done by a force of 1N on an object that displaces it through a distance of 1m in the direction of the applied force.

Question 4.
A pair of bullocks exerts a force of 140N on the plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?
Answer:
Work done by the bullocks is given by the expression:
Work done = Force × Displacement
W=F × s
Where,
Applied force, F = 140N
Displacement, s = 15 m
W= 140 × 15 = 2100J
Hence, 2100 J of work is done in ploughing the length of the field.

Page 152

Question 1.
What is the kinetic energy of an object?
Answer:
The energy possessed by a body by the virtue of its motion is called kinetic energy. It has various applications:
(a) Kinetic energy of flowing water is used to generate electricity.
(b) Kinetic energy of hammer is used in driving a nail into a log of wood.
(c) Kinetic energy of air is used to run windmills, etc.

Question 2.
Write an expression for the kinetic energy of an object.
Answer:
If a body of mass ‘m’ is moving with a velocity ‘v’ its kinetic energy K.E. is given by the expression,
K.E. = \(\frac{1}{2}\) mv²
Its SI unit is joule (J).

Question 3.
The kinetic energy of an object of mass, m moving with a velocity of 5 m s^-1 is 25J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
K.E. of the object = 25J
Velocity of the object, v = 5 m/s
∴ K.E. = \(\frac{1}{2}\) mv²
25 = \(\frac{1}{2}\)m(5)²
m = 2 kg

(a) If velocity is doubled, v = 10m/s, m = 2 kg
K.E. = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\)(2) × (10)² = 100J.
K.E will become four times.

(b) If velocity is increased 3 times,
v = 15 m/s, m = 2 kg
K.E = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) × 2 × (15)² = 225J.
K.E will become nine times.

Page 156

Question 1.
What is power?
Answer:
Power is the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by the expression,
Power = \(\frac{Work}{Time }\) = \(\frac{w}{t}\)
It is expressed in watt (W).
watt = \(\frac{Joule}{Second}\)

Question 2.
Define 1 watt of power.
Answer:
A body is said to have a power of 1 watt if it does 1 joule work in 1 s, i.e.,
1W =\(\frac{1 \mathrm{~J}}{1 \mathrm{~s}}\)

Question 3.
A lamp consumes 1000J of electrical energy in 10s. What is its power?
Answer:
Power = \(\frac{Work done}{Time }\)
Work done = Energy consumed by the lamp = 1000J
Time = 10s
Power = \(\frac{1000}{10}\) = 100J s^-1 = 100W

Question 4.
Define average power.
Answer:
The average power is defined as the total work done or total energy consumed divided by the total time taken.
Average power = \(\frac{Total work done}{Total time taken }\)

JAC Class 9th Science Work and Energy Textbook Questions and Answers

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
1. Suma is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A windmill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Foodgrains are getting dried in the sun.
7. A sailboat is moving due to wind energy.
Answer:
Work is done whenever the following two conditions are satisfied:

• A force acts on the body.
• There is a displacement of the body by the application of force.

1. While swimming, Suma applies a force to push the water backwards. Therefore, Suma swims in the forward direction under the influence of the forward reaction of water. Here, the force causes a displacement. Hence, work is done by Suma while swimming.
2. While carrying a load, the donkey has to apply a force in the upward direction. But, displacement of the load is in the forward direction. Since, displacement is perpendicular to force, the work done is zero.
3. A windmill works against the gravitational force to lift water. Hence, work is done by the windmill in lifting water from the well.
4. In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero.
5. An engine applies force to pull the train. This allows the train to move in the direction of force. Therefore, there is a displacement of the train in the direction of force. Hence, work is done by the engine on the train.
6. Foodgrains do not move in the presence of solar energy. Hence, the work done is zero when the foodgrains are getting dried in the sun.
7. Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

Question 2.
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Work done by the force of gravity on the object is zero. Force of gravity acts in the vertically downward direction and the distance covered by the object is in the horizontal direction. As there is no  displacement in the direction of force, hence the work done is zero.

Question 3.
A battery lights a bulb Describe the energy changes involved in the process.
Answer:
When a bulb is connected to a battery, the chemical energy of the battery is transformed into electrical energy. When the bulb receives this electrical energy, it converts it into heat and light energy. Hence, the transformation of energy in the given situation can be shown as:
Chemical Energy → Electrical Energy → Heat Energy + Light Energy

Question 4.
Certain force acting on a 20 kg mass changes its velocity from 5 ms^-1 to 2 ms^-1. Calculate the work force.
Answer:
Here,
m = 20 kg,
u = 5 m^-1,
v = 2 m^-1
Work done = Change in K.E.
W = \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mu²
= \(\frac{1}{2}\) × 20 × (2² – 5²)
= 10 × (4 – 25)
= – 10 × 21 = – 210J
The negative sign indicates that work is done against the applied force.

Question 5.
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Therefore, work done by gravity is given by the expression,
W = mgh
Where, vertical displacement, h = 0
∴ W = mg × 0 = 0
Hence, the work done by gravity on the body is zero.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
No. The process does not violate the law of conservation of energy. This is because when a body falls from a height, its potential energy progressively changes into kinetic energy. The decrease in potential energy is equal to the increase in kinetic energy of the body. During this process, the total mechanical energy of the body remains conserved. Therefore, the law of conservation of energy is not violated.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:
While riding a bicycle, the muscular energy of the rider gets transformed into heat energy and kinetic energy of the bicycle. Heat energy heats the rider’s body. Kinetic energy provides velocity to the bicycle. The transformation can be shown as: Muscular Energy → Kinetic Energy + Heat Energy During this transformation, the total energy remains conserved.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
When we push a huge rock, there is no transfer of muscular energy to the stationary rock. Also, there is no loss of energy because muscular energy is transferred into heat energy which causes our body to become hot.

Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer:
1 unit of energy is equal to 1 kilowatt hour (kW h).
1 unit = 1 kW h
1 kWh = 3.6 × 10⁶J
Therefore, 250 units of energy
= 250 × 3.6 × 10⁶ = 9 × 10⁸J

Question 10.
An object of mass 40 kg is raised to a height of 5m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half – way down.
Answer:
Gravitational potential energy is given by the expression,
P_E= mgh Where,
h = Vertical displacement = 5 m
m = Mass of the object = 40 kg
g = Acceleration due to gravity = 9.8 ms^-2
P_E = 40 × 5 × 9.8= 1960J.
At half – way down, the potential energy of the object will be \(\frac{1960}{2}\) = 980J.
At this point, the object has an equal amount of potential and kinetic energy. This is due to the law of conservation of energy. Hence, half – way down, the kinetic energy of the object will also be 980J.

Question 11.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:
Zero. When the satellite moves around the earth, the force of gravity acts on it along the radius of its orbit, while its direction of motion is along the tangent to the orbit at any point. Thus, the force acts perpendicular to the displacement. Hence, the work done on the satellite is zero.

Question 12.
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Answer:
Yes. For a uniformly moving object. Suppose an object is moving with a constant velocity. The net force acting on it is zero. But, there is a displacement along the motion of the object. Hence, there can be a displacement without a force.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
When a person holds a bundle of hay over his head, there is no displacement in the bundle of hay. Although, force of gravity is acting on the bundle, the person is not applying any force on it. Hence, in the absence of force and displacement, work done by the person on the bundle is zero.

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Ans.
Energy consumed by an electric heater can be obtained with the help of the expression,
P= \(\frac{\mathrm{W}}{\mathrm{t}}\)
Where, power rating of the heater,
P= 1500 W
Time for which the heater has operated, t = 10h
Work done = Energy consumed by the heater
Therefore, energy consumed
= P × t = 1500 W × 10 h
= \( \frac{15000}{1000}\) = 15kW h

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Answer:
The law of conservation of energy states that energy can neither be created nor destroyed. It can only be converted from one form to another. Consider the case of an oscillating pendulum. When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height ‘h’ above the mean level ‘P’. At this point, the kinetic energy of the bob changes completely into potential energy.

The kinetic energy becomes zero, and the bob possesses only potential energy. As it moves towards point P, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates. The bob does not oscillate forever. It comes to rest because air resistance resists its motion.

The pendulum loses its kinetic energy to overcome this friction and stops after some time. The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system remains conserved.

Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer:
Work done on the object = Change in K.E. of the object.
W = \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) m(0)² = \(\frac{1}{2}\) mv²
To bring the object to rest, \(\frac{1}{2}\) mv² amount of work is required to be done on the object.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Kinetic energy, KE = \(\frac{1}{2}\) mv²
Where,
mass of the car, m = 1500 kg and velocity of the car,
v = 60 km/h = \(\frac{60 \times 5}{18}\) ms^-1
KE = \(\frac{1}{2}\) × 1500 × \(\left(\frac{60 \times 5}{18}\right)^{2}\)
= 20.8 × 10⁴ J
Hence, 20.8 × 10⁴ J of work is required to stop the car.

Question 18.
In each of the following, a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the . force is negative, positive or zero.

Answer:

1. In this case, the direction of force acting on the block is perpendicular to the displacement. Therefore, work done by force on the block will be zero.
2. In this case, the direction of force acting on the block is in the direction of displacement. Therefore, work done by force on the block will be positive.
3. In this case, the direction of force acting on the block is opposite to the direction of displacement. Therefore, work done by the force on the block will be negative.

Question 19.
Soni says that the acceleration of an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
Acceleration in an object could be zero even when several forces are acting on it. This happens when all the forces cancel out each other, i.e., the net force acting on the object is zero. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

Question 20.
Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Answer:
Here, P = 500 W = \(\frac{500}{1000}\) kW= 0.5 kW and t =10 h
Energy consumed by four devices
= 4P × t = 4 × 0.5 kW × 10h = 20 kW h

Question 21.
A freely falling object eventually stops reaching the ground. What happens to its kinetic energy?
Answer:
When a body is falling freely, the energy remains conserved. Potential energy is transformed into kinetic energy. But when the body reaches the ground, its kinetic energy is converted into sound energy and also into a little heat energy (due to friction between ground and object). A part of the energy is also transferred to the ground. Hence, the total energy of the system remains conserved.

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter 13 Why Do We Fall Ill

JAC Class 9th Science Why Do We Fall Ill InText Questions and Answers

Page 178

Question 1.
State any two conditions essential for good health.
Answer:
Two conditions essential for good health are.
(a) An individual must have better health facilities and more professionals around to deal with health problems.
(b) All basic necessary conditions to prevent diseases must be present. For example, proper garbage collection and disposal, cleaning of drains, supply of clean drinking water, etc.

Question 2.
State any two conditions essential for being free of disease.
Answer:
The two conditions essential for being free of disease are.
(a) Personal hygiene and cleanliness are necessary to stay away from diseases.

(b) Individuals should take a balanced diet that contains carbohydrates, fats, proteins, vitamins, fibres, and adequate quantity of water.

Question 3.
Are the answers to the above questions necessarily the same or different? Why?
Answer:
The answers to the above questions are different because a person may be free of disease but his mental, social or economical health may not be good.

Page 180

Question 1.
List any three reasons why you would think that you are sick and ought to see a doctor. If only one of these symptoms were present, would you still go to the doctor? Why or why not?
Answer:
The three reasons are:
(a) High fever
(b) Diarrhoea
(c) Stomach pain
The above symptoms basically indicate that there might be a disease, but we cannot predict the kind of disease. Therefore, it becomes necessary to visit a doctor so that the disease can be identified and can be treated with proper medication. However, if any of these symptoms are present, we usually do not visit a doctor. This is because such symptoms do not have much effect on our general health and ability to work. However, if a person is experiencing these symptoms for quite some time, he needs to visit a doctor for proper treatment.

Question 2.
In which of the following cases do you think the long-term effects on your health are likely to be most unpleasant?

• if you get jaundice,
• if you get lice,
• if you get acne. Why?

Answer:
The long-term effects on our health are likely to be most unpleasant if we get jaundice because its symptoms severely affect our internal organs and persist for a long time.

Page 186

Question 1.
Why are we normally advised to take bland and nourishing food when we are sick?
Answer:
We are normally advised to take bland and nourishing food when we are sick because we are weak during illness and need complete nourishment for faster recovery.

Question 2.
What are the different means by which infectious diseases are spread?
Answer:
The different means by which infectious diseases spread are.
(a) Air. They are also called air-borne diseases. The air carries disease causing bacteria, viruses, etc., and the diseases that can be caused are. common cold, influenza, tuberculosis, etc

(b) Food and water: When one eats or drinks contaminated food/water that contains bacteria, viruses, worms, etc., it can cause diseases like cholera, typhoid, hepatitis, etc.

(c) Contact: Many diseases spread by contact of infected person with the healthy person, e.g., fungal infections, skin diseases, scabies, etc.

(d) Sexual contact: Many diseases can be transmitted by sexual contact, e. g., syphilis, AIDS, etc.

(e) Body fluids: Fluids like blood, semen and mother’s milk, when infected, can also cause diseases., e.g., AIDS.

(f) Vectors: The organism that spreads a disease by carrying pathogens from one place to another is called vector, e.g., mosquitoes are vectors that carry pathogens like protozoa which can cause malaria.

Question 3.
What precautions can you take in your school to reduce the incidence of infectious diseases?
Answer:
The precautions that one can take in school to reduce the incidence of infectious diseases are as follows.

1. Using handkerchief while coughing or sneezing.
2. Washing hands before eating food.
3. Staying at home if anyone suffers from infectious diseases.
4. Getting vaccinated well before the infection affects.
5. Keeping the school surroundings clean and checking for stagnant water.

Question 4.
What is immunization?
Answer:
Immunization is a method of preparing memory cells in immune system. In this method, weak or dead microbes of a disease are injected in the body such that they develop memory lymphocytes (WBCs) but do not cause the actual disease, e.g., vaccination against polio, tetanus, etc.

Question 5.
What are the immunization programmes available at the nearest health centre in your locality? Which of these diseases are the major health problems in your area?
Answer:

1. DPT vaccine for diphtheria, pertussis and tetanus in infants.
2. Pulse polio vaccine to prevent polio.
3. BCG vaccine for tuberculosis.
4. Hepatitis vaccine for hepatitis.
5. TAB vaccine for typhoid. Tuberculosis and diarrhoea are the major health problems in our area.

JAC Class 9th Science Why Do We Fall Ill Textbook Questions and Answers

Question 1.
How many times did you fall ill in the last one year? What were the illnesses?
(a) Think of one change you could make in your habits in order to avoid any of / most of the above illnesses.
(b) Think of one change you would wish for in your surroundings in order to avoid any of / most of the above illnesses.
Answer:
This varies from person to person. Some people fall ill several times in a year while others do not fall ill at all. A person’s immune system and hygiene-related habits play a major role in determining the person’s health.

Question 2.
A doctor/nurse/health worker is exposed to more sick people than others in the community. Find out how she/he avoids getting sick herself/ himself?
Answer:
Doctors/nurses/health workers when exposed to sick people keep their nose and mouth covered, take care of hygiene, wash hands with soap before drinking water or eating food. They use mask, gloves, etc., to avoid direct contact with the person suffering from infectious diseases.

Question 3.
Conduct a survey in your neighbourhood to find out what the three most common diseases are. Suggest three steps that could be taken by your local authorities to bring down the incidence of these diseases.
Answer:

• Three most common diseases are:
  1. (a) tuberculosis
  2. (b) typhoid
  3. (c) jaundice
• Steps to be taken to bring down the incidence of these diseases are:
  1. (a) Proper disposal of sewage.
  2. (b) Ensuring supply of safe drinking water.
  3. (c) Providing a clean environment and preventing mosquitoes from breeding.

Question 4.
A baby is not able to tell his/her caretakers that she/he is sick. What would help us to find out.
(a) that the baby is sick?
(b) what is the sickness?
Answer:
(a) We can find out that the baby is sick by his/her behavioural changes such as constant crying, improper intake of food, frequent mood change, etc.

(b) The sickness is determined by symptoms or indications that can be seen in the baby. The symptoms may include vomiting, fever, loose motions, paleness in the body, etc.

Question 5.
Under which of the following conditions is a person most likely to fall sick?
(a) When she is recovering from malaria.
(b) When she has recovered from malaria and is taking care of someone suffering from chickenpox.
(c) When she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chickenpox. Why?
Answer:
A person is more likely to fall sick when she is on a four-day fast after recovering from malaria and is taking care of someone who is suffering from chickenpox. This is because she is fasting during recovery and her immune system is so weak that it is not able to protect its own body from any foreign infection. If she is taking care of someone suffering from chickenpox, she has more chances of getting infected by the chickenpox virus and will get sick again with this disease.

Question 6.
Under which of the following conditions are you most likely to fall sick? Give reasons.
(a) When you are taking examinations.
(b) When you have travelled by bus and train for two days.
(c) When your friend is suffering from measles.
Answer:
You are more likely to fall sick when your friend is suffering from measles. This is because measles is highly contagious and can easily spread through breathing, i. e., through air. Thus, if your friend is suffering from measles, take proper care otherwise you might easily get infected with the disease.

JAC Class 9 Science Solutions

JAC Jharkhand Board Class 10th English Solutions First Flight & Footprints without Feet

JAC Board Class 10th English Solutions First Flight

JAC Jharkhand Board Class 10th English First Flight Prose

• Chapter 1 A Letter to God
• Chapter 2 Nelson Mandela: Long Walk to Freedom
• Chapter 3 Two Stories about Flying
  • Chapter 3(i) His First Flight
  • Chapter 3(ii) The Black Aeroplane
• Chapter 4 From the Diary of Anne Frank
• Chapter 5 The Hundred Dresses Part 1
• Chapter 6 The Hundred Dresses Part 2
• Chapter 7 Glimpses of India
  • Chapter 7(i) A Baker From Goa
  • Chapter 7(ii) Coorg
  • Chapter 7(iii) Tea From Assam
• Chapter 8 Mijbil the Otter
• Chapter 9 Madam Rides the Bus
• Chapter 10 The Sermon at Benares
• Chapter 11 The Proposal

JAC Jharkhand Board Class 10th English First Flight Poem

• Poem 1 Dust of Snow
• Poem 2 Fire and Ice
• Poem 3 A Tiger in the Zoo
• Poem 4 How to Tell Wild Animals
• Poem 5 The Ball Poem
• Poem 6 Amanda
• Poem 7 Animals
• Poem 8 The Trees
• Poem 9 Fog
• Poem 10 The Tale of Custard the Dragon
• Poem 11 For Anne Gregory

JAC Class 10th English Solutions Footprints without Feet

JAC Jharkhand Board Class 10th English Supplementary Reader Footprints without Feet

• Chapter 1 A Triumph of Surgery
• Chapter 2 The Thief’s Story
• Chapter 3 The Midnight Visitor
• Chapter 4 A Question of Trust
• Chapter 5 Footprints without Feet
• Chapter 6 The Making of a Scientist
• Chapter 7 The Necklace
• Chapter 8 The Hack Driver
• Chapter 9 Bholi
• Chapter 10 The Book that Saved the Earth

JAC Class 10 English Reading Comprehension

• Discursive Passages
• Factual Passages

JAC Class 10 English Grammar

• Tenses
• Modals
• Subject-Verb Concord
• Reported Speech
• Determiners

JAC Class 10 English Writing

• Letter Writing
• Application Writing
• E-mail Writing
• Analytical Paragraph Writing

JAC Board Class 9th Science Solutions Chapter 12 Sound

JAC Class 9th Science Sound InText Questions and Answers

Poge 162

Question 1.
How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
When an object vibrates, it sets the particles of the medium around it in vibration. The particles in the medium in contact with the vibrating object are displaced from their equilibrium position. It then exerts a force on the adjacent particles. After displacing the adjacent particle, the first particle of the medium comes back to its original position. This process continues in the medium till the sound reaches our ear.

Page 163

Question 1.
Explain how sound is produced by your school bell.
Answer:
When the bell continues to vibrate forward and backward, it creates a series of compressions and rarefactions resulting in the production of sound.

Question 2.
Why are sound waves called mechanical waves?
Answer:
Sound waves need material medium to propagate, therefore, they are called mechanical waves. Sound waves can propagate through a medium only because of the interaction of the particles present in that medium.

Question 3.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
No, because a sound wave needs a medium through which it can propagate. Since there is no atmosphere on the moon, no material medium is available and we cannot hear any sound on the moon.

Page 166

Question 1.
Which wave property determines:
(a) loudness
(b) pitch?
Answer:
(a) Amplitude
(b) Frequency

Question 2.
Guess which sound has a higher pitch: guitar or car horn?
Answer:
Guitar has a higher pitch than car horn, because sound produced by the strings of guitar has a higher frequency than that of a car horn. The higher the frequency, the higher is the pitch.

Page 166

Question 1.
What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:

1. Wavelength: The distance between two consecutive compressions or two consecutive rarefactions is known as wavelength. Its SI unit is metre (m).
2. Frequency: The number of complete oscillations per second is known as the frequency of a sound wave. It is measured in hertz (Hz).
3. Time period: The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the wave.
4. Amplitude: The maximum height reached by the crest or the maximum depth reached by the trough of a
   sound wave is called its amplitude.

Question 2.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
Speed, wavelength and frequency of a sound wave are related by the following equation:
Speed (v) = Wavelength (λ) x Frequency (v)
V = λ × v

Question 3.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:
Frequency of the sound wave, v = 220 Hz
Speed of the sound wave, v = 440 ms^-1
For a sound wave,
Speed = Wavelength × Frequency
v = λ × v
∴ λ = \(\frac{v}{v}\) = \(\frac{440}{220}\) = 2
Hence, the wavelength of the sound wave is 2 m.

Question 4.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Answer:
The time interval between two successive compressions is equal to the time period of the wave. This time period is reciprocal of the frequency of the wave and is given by the relation:
T = \( \frac{1}{\text { Frequency }}\) = \(\frac{1}{500}\) = 0.002 s

Page 166

Question 1.
Distinguish between loudness and intensity of sound.
Answer:
Intensity of a sound wave is defined as the amount of sound energy passing through a unit area per second. Loudness is a measure of the response of the ear to the sound. The loudness of a sound is defined by its amplitude.

Page 167

Question 1.
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
The speed of sound depends on the nature of the medium. Sound travels the fastest in solids. Its speed decreases in liquids and it is the slowest in gases. Therefore, for a given temperature, sound travels fastest in iron.

Page 168

Question 1.
An echo is heard in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms^-1?
Answer:
Speed of sound, v = 342 ms^-1
Echo returns in time, t = 3s
Distance travelled by sound
= v × t = 342 × 3 = 1026 m
In the given time interval, sound has to travel a distance that is twice the distance between the reflecting surface and the source.
Hence, the distance of the reflecting surface from the source
= \(\frac{1026}{2}\) m = 513 m.

Page 169

Question 1.
Why are the ceilings of concert halls curved?
Answer:
The ceilings of concert halls are curved so that the sound, after reflection (from the walls), spreads uniformly in all directions.

Page 170

Question 1.
What is the audible range of the average human ear?
Answer:
The audible range of an average human ear is 20 Hz to 20,000 Hz.

Question 2.
What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
Answer:
(a) Infrasound has frequencies less than 20 Hz.
(b) Ultrasound has frequencies more than 20,000 Hz.

Page 172

Question 1.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Answer:
Time taken by the sonar pulse to return, t = 1.02s
Speed of sound in salt water, v = 1531 ms^-1
Distance travelled by the sonar pulse
= Speed of sound × Time taken
= 1.02 × 1531 = 1561.62 m
Distance travelled by the sonar pulse during its transmission and reception in water
= 2 × actual distance = 2d
Actual distance of the cliff from the submarine, d
Distance travelled by the sonar pulse
= \(\frac{Distance travelled by the sonar pulse2}{2}\) = \( \frac{1561.62}{2}\) = 780.81 m

JAC Class 9th Science Sound Textbook Questions and Answers

Question 1.
What is sound and how is it produced?
Answer:
Sound is a form of energy which gives the sensation of hearing. It is produced by the vibrations caused in the medium by vibrating objects.

Question 2.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:
When a vibrating body moves forward, it creates a region of high pressure in its vicinity. This region of high pressure is known as compression. When it moves backward, it creates a region of low pres – sure in its vicinity. This region is known as rarefaction. As the body continues to move forward and backward, it produces a series of compressions and rarefactions. This is shown in the figure below.

Question 3.
Cite an experiment to show’ that sound needs a material medium for its propagation.
Answer:
Take an electric bell and an air tight glass bell jar connected to a vacuum pump. Suspend the bell inside the jar, and press the switch of the bell. You will be able to hear the bell ring. Now pump out the air from the glass jar. The sound of the bell will become progressively fainter and after some time, the sound will not be heard. This is so because almost all air has been pumped out. This shows that sound needs a material medium to travel.

Question 4.
Why is sound wave called a longitudinal wave?
Answer:
Sound wave is called a longitudinal wave because it is produced by compressions and rarefactions in the air. The air particles vibrate parallel to the direction of propagation of sound.

Question 5.
Which characteristics of the sound help you to identify your friend by his voice while sitting with others in a dark room?
Answer:
The quality or timber of sound enables us to identify our friend by his voice.

Question 6.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer:
The speed of sound (344 m/s) is less than the speed of light (3 × 108 m/s). Sound of thunder takes more time to reach the earth as compared to light. Hence, a flash is seen before we hear a thunder.

Question 7.
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms^-1.
Answer:
For a sound wave,
Speed = Wavelength × Frequency = λ × v
Speed of sound in air = 344 m/s (Given)
(a) For, v = 20 Hz
λ₁ = \(\frac{v}{v}\) = \(\frac{344}{20}\)
= 17.2 m

(b) For, v = 20000 Hz
λ₂ = \(\frac{v}{v}\) = \(\frac{344}{20,000}\)
= 0.0172 m
Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

Question 8.
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Answer:
Velocity of sound in air = 346 m/s
Velocity of sound wave in aluminium = 6420 m/s
Let the length of the rod be l
Time taken for sound wave in air,
t₁ =\(\frac{l}{Velocity in air}\)
Velocity in air Time taken for sound wave in aluminium,
t₂ = \(\frac{l}{Velocity in aluminium}\)
Therefore, = \(\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}\)
= \(\frac{Velocity in aluminium}{Velocity in air}\) = \(\frac{6420}{ 346}\)
= 18.55

Question 9.
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Frequency = 100 Hz (given)
This means that the source of sound vibrates 100 times in one second. Therefore, number of vibrations in 1 minute, i.e., in 60 seconds = 100 × 60 = 6000 times.

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:
Sound follows the same laws of reflection as light does. The incident sound wave and the reflected sound wave make equal angles with the normal to the surface at the point of incidence. Also, the incident sound wave, the reflected sound wave and the normal to the point of incidence, all lie in the same plane.

Question 11.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
Answer:
An echo is heard when the time for the reflected sound is heard after 0.1s.
Time taken = \(\frac{Total distance}{Velocity}\)
On a hotter day, the velocity of sound is more. If the time taken by echo is less than 0.1s, it will not be heard.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
Two practical applications of reflection of sound waves are:

1. Reflection of sound is used to measure the distance and speed of underwater objects. This technique is known as SONAR.
2. Working of a stethoscope is also based on reflection of sound. In a stethoscope, the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflections.

Question 13.
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g =10 ms^-2 and speed of sound = 340 ms^-1.
Answer:
Height of the tower, s = 500m
Velocity of sound, v = 340 ms^-1.
Acceleration due to gravity, g = 10 ms^-2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Let the time taken by the stone to fall to the base of the tower be t₁
According to the second equation of motion:
s= ut₁ + \(\frac{1}{2} \mathrm{gt}_{1}^{2}\)
500 = (0 × t₁) + \( \left(\frac{1}{2} \times 10 \times t_{1}^{2}\right)\)
\( \mathrm{t}_{1}^{2}\) = 100
t₁ = 10s
Now, time taken by the sound to reach the top from the base of the tower, t₂ = \(\frac{500}{340}\) = 1.47s
Therefore, the splash is heard at the top after time, t.
Where, t = t₁ + t₂ = 10 + 1.47 = 11.47s.

Question 14.
A sound wave travels at a speed of 339 ms^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer:
Speed of sound, v = 339 ms^-1
Wavelength of sound,
λ = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
v = λ × v
∴ v = \(\frac{v}{λ}\) = \(\frac{ 339 }{0.015}\) = 22600 Hz
The frequency range of audible sound for humans is between 20 Hz to 20,000 Hz. Since the frequency of the given sound is more than 20,000 Hz, it is not audible.

Question 15.
What is reverberation? How can it be reduced?
Answer:
The repeated or multiple reflections of sound in a large enclosed space is known as reverberation. The reverberation can be reduced by covering the ceiling and walls of the enclosed space with sound absorbing materials, such as fibre board, loose woollens, etc.

Question 16.
What is loudness of sound? What factors does it depend on?
Answer:
Loudness is a measure of sound energy reaching the ear per second. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Question 17.
Explain how bats use ultrasound to catch a prey.
Answer:
Bats produce high – pitched ultrasonic squeaks. These high – pitched squeaks are reflected by objects and their preys and returned to the bats’ ears. This allows the bat to know the distance and direction of their prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Objects to be cleaned are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasonic waves detaches the dirt from the objects.

Question 19.
Explain the working and application of a sonar.
Answer:
SONAR is an acronym for sound Navigation And Ranging. It is an acoustic device used to measure the depth, direction and speed of underwater objects, such as submarines and ship wrecks, with the help of ultrasounds. It is also used to measure the depth of seas and oceans.

A beam of ultrasonic sound is produced and transmitted by the transducer (a device that produces ultrasonic sound) of the SONAR, which travels through sea water.
The echo produced by the reflection of this ultrasonic sound is detected and recorded by the detector, which is converted into electrical signals. The distance (d) of the underwater object is calculated from the time (t) taken by the echo to return with speed (v) which is given by 2d = v × t.
This method of measuring distance is also known as ‘echo – ranging’.

Question 20.
A sonar device on a submarine sends out a signal and receives an echo 5s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625m.
Time taken to hear the echo, t = 5s
Distance of the object from the submarine, d = 3625m
Total distance travelled by the sonar waves during the transmission and reception in water = 2d
Velocity of sound in water,
v = \(\frac{2 \mathrm{~d}}{t}\) = \(\frac{2 \times 3625}{5}\) = 1450ms^-1

Question 21.
Explain how7 defects in a metal block can be detected using ultrasound.
Answer:
Defects in metal blocks do not allow ultrasound waves to pass through them and they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. The defective part of the metal block does not allow ultrasound to pass through it. As a result, it will not be detected by the detector. Hence, defects in metal blocks can be detected using ultrasound.

Question 22.
Explain how the human ear works.
Answer:
The human ear consists of three parts – outer ear, middle ear and inner ear.

1. Outer ear: This is also called ‘pinna’. It collects the sound from the surroundings and directs it towards auditory canal.
2. Middle ear: The sound reaches the end of the auditory canal where there is a thin membrane called eardrum or tympanic membrane. The sound waves set this membrane in vibration. These vibrations are amplified by three small bones. Malleus, Incus and Stapes in the middle ear.
3. Inner ear: These vibrations reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound.
   

JAC Class 9 Science Solutions

JAC Board Class 9th Science Solutions Chapter Chapter 15 Improvement in Food Resources

JAC Class 9th Science Improvement in Food Resources InText Questions and Answers

Page 204

Question 1.
What do we get from cereals, pulses, fruits and vegetables?
Answer:
Cereals provide us carbohydrates which give energy. Pulses provide us proteins which build our body. Vegetables and fruits provide vitamins and minerals.

Page 205

Question 1.
How do biotic and abiotic factors affect crop production?
Answer:

1. Factors responsible for loss of grains, during storage and production are:
   • Biotic factors like rodents, pests, insects, etc.
   • Abiotic factors like temperature, humidity, moisture, etc.
2. Combination of both biotic and abiotic factors causes:
   • infestation of insects
   • weight loss
   • poor germination ability
   • degradation in quality
   • discolouration
   • poor market price

Question 2.
What are the desirable agronomic characteristics for crop improvements?
Answer:
Desirable agronomic characteristics for crop improvements are as follows:

1. Tallness and profuse branching are desirable characters for fodder crops.
2. Dwarfness is desired in cereals so that less nutrients are consumed by these crops.

Page 206

Question 1.
What are macronutrients and why are they called macronutrients?
Answer:
Macronutrients are the essential elements which are utilised by plants in large quantities. Many macronutrients are required by the plants for the following functions:

1. As the constituent of protoplasm
2. N, P, S are present in proteins
3. Ca is present in cell wall
4. Mg is an important constituent of chlorophyll

Question 2.
How do plants get nutrients?
Answer:
Plants get nutrients from air, water and soil. There are sixteen nutrients essential for the growth of plants. Carbon and oxygen are supplied by air. Hydrogen is supplied by water. The remaining thirteen nutrients are supplied by soil.

Page 207

Question 1.
Compare the use of manures and fertilisers in maintaining soil fertility.
Answer:

1. Effects of using manures on soil quality:
   • The manures enrich the soil with nutrients.
   • They provide a lot of organic matter (humus) to the soil and thus restore water retention capacity of sandy soils and drainage in clayey soils.
   • The addition of manures reduces soil erosion.
   • They provide food for soil organisms, like soil friendly bacteria.
2. Effects of using fertilisers on soil quality:
   • By the continuous use of fertilisers, the soil becomes powdery, dry and the rate of soil erosion increases.
   • By the use of fertilisers, the organic matter decreases which further decrease the porosity of soil and the plant roots do not get oxygen properly.
   • The nature of soil changes to acidic or basic.

Page 208

Question 1.
Which of the following conditions will give the most benefits? Why?
(a) Farmers use high-quality seeds, do not adopt irrigation or use fertilisers.
(b) Farmers use ordinary seeds, adopt irrigation, use fertilisers.
(c) Farmers use quality seeds, adopt irrigation, use fertilisers and use crop protection measures.
Answer:
Use of any quality seeds is not sufficient until they are properly irrigated, enriched with fertilisers and protected from biotic factors. Hence, option (c) will give the most benefits.

Page 209

Question 1.
Why should preventive measures and biological control methods be preferred for protecting crops?
Answer:
Diseases in plants are caused by pathogens. To get rid of pathogens, some preventive measures and biological control methods are used as they are simple, economical and minimise pollution without affecting the soil quality.

Question 2.
What factors may be responsible for losses of grains during storage?
Answer:
The factors responsible for the losses of grains during storage are:

1. Abiotic factors like moisture (present in food grains), humidity (of air) and temperature.
2. Biotic factors like insects, rodents, birds, mites, bacteria and fungi.

Page 210

Question 1.
Which method is commonly used for improving cattle breeds and why?
Answer:
Cross breeding is a process in which indigenous varieties of cattle are crossed by exotic breeds to get a breed which is high yielding. During cross – breeding, the desired characters are taken into consideration. The offspring should be high yielding, should have early maturity and should be resistant to climatic conditions.

Page 211

Question 1.
Discuss the implications of the following statement: “It is interesting to note that poultry is India’s most efficient converter of low fibre foodstuff (which is unfit for human consumption) into highly nutritious animal protein food”.
Answer:
The basic aim of poultry farming is to raise domestic fowl for egg production and chicken for meat. These poultry birds are not only the efficient converters of agricultural by – products, particularly cheaper fibrous wastes (which are unfit for human consumption but can be formulated into cheaper diets for poultry birds) into high – quality meat but also help in providing eggs, feathers and nutrient – rich manure. For this reason, it is said that, “poultry is India’s most efficient converter of low fibre foodstuff into highly nutritious animal protein food”.

Page 211

Question 1.
What management practices are common in dairy and poultry farming?
Answer:
Management practices are common in dairy and poultry farming:

1. Shelter: Dairy animals and poultry birds require proper shelter, i.e., well – designed, dairy and hygienic shelter.
2. Feeding: To get good yield of food production, proper feed is provided to dairy animals and poultry birds.
3. Caring for animal health: Animals and birds must be protected from diseases caused by viruses, bacteria or fungi.

Question 2.
What are the differences between broilers and layers and in their management?
Answer:
The poultry bird groomed for obtaining meat is called broiler. The egg laying poultry bird is called layer. The housing, nutritional and environmental requirements of broilers are somewhat different from those of egg layers. The ration (daily food requirement) for broilers is protein- rich with adequate fat. The level of vitamins A and K is kept high in the poultry feeds. Layers require enough space and proper lightning. They are vaccinated against various diseases. Broilers are taken care of to avoid mortality and to maintain feathering and carcass quality.

Page 213

Question 1.
How are fish obtained?
Answer:
There are two ways of obtaining fish. One is from natural resources, which is called capture fishing. The other way is by fish farming, which is called culture fishery.

Question 2.
What are the advantages of composite fish culture?
Answer:
In composite fish culture, a combination of five or six fish species is cultured in a single fish pond. These species are selected so that they do not compete for food among themselves and have different types of food habits. As a result, the food available in all the parts of the pond is used. For example, Catlas are surface feeders, Rohus feed in the middle – zone of the pond, Mrigals and common carps are bottom feeders and grass carps feed on the weeds. Together, these species can use all the food in the pond without competing with each other. This increases the fish yield from the pond.

Page 213

Question 1.
What are the desirable characters of bee varieties suitable for honey production?
Answer:
The desirable characters of bee varieties suitable for honey production:

1. The variety of bee should be able to collect a large amount of honey.
2. The bees should stay in a given beehive for a longer period.
3. The bees should have capacity of breeding well.
4. The variety of bee should be disease resistant.

Question 2.
What is pasturage and how is it related to honey production?
Answer:
Pasturage means the flowers available to the bees for nectar and pollen collection. In addition to adequate quantities of pasturage, the kind of flowers available will determine the taste of the honey.

JAC Class 9th Science Improvement in Food Resources Textbook Questions and Answers

Question 1.
Explain any one method of crop production which ensures high yield.
Answer:
One method used for crop production which ensures high yield is plant breeding. It is the science involved in improving the varieties of crops by breeding plants. The plants from different areas/places are picked up with desired traits and then hybridisation or cross-breeding of these varieties is done to obtain a plant/crop of desired characteristics. The high – yielding crop variety shows the following characteristics. High yield, early maturation, less water for irrigation, better quality seeds are produced, less fertiliser required, adapts itself to the environmental conditions.

Question 2.
Why are manure and fertilisers used in fields?
Answer:
Manures and fertilisers are used in fields to enrich the soil with the required nutrients, to improve fertility of the soil, crop yield and quality. Manure helps in enriching the soil with organic matter and nutrients. This improves the fertility and structure of the soil. On the other hand, fertilisers ensure a healthy growth and development in plants. They are a good source of nitrogen, phosphorus and potassium. A balanced combination of manures and fertilisers in the soil should be used to get an optimum yield.

Question 3.
What are the advantages of intercropping and crop rotation?
Answer:
Both Intercropping and crop rotation are used to get maximum benefit on limited land:

1. Intercropping ensures maximum utilisation of the nutrients supplied by the soil. It helps in preventing pests and diseases to spread throughout the field. It also increases soil fertility.
2. Crop rotation prevents soil depletion, increases soil fertility, and reduces soil erosion. Both these methods reduce the need for fertilisers. These also help in controlling weeds and control the growth of pathogens and pests in the crops.

Question 4.
What is genetic manipulation? How is it useful in agricultural practices?
Answer:
Genetic manipulation is a process of incorporating desirable (genes) characters into crop varieties by hybridisation. Hybridisation involves crossing between genetically dissimilar plants. This is done for production of varieties with desirable characteristics like profuse branching in fodder crops, high – yielding varieties in maize, wheat, etc.
Genetic manipulation is useful in developing varieties which show:

1. increased yield
2. better quality
3. shorter and early maturity period
4. better adaptability to adverse environmental conditions
5. desirable characteristics

Question 5.
How do storage grain losses occur?
Answer:
The storage grain losses can occur due to the following reasons:

1. Biotic factors: Like insects, rodents, fungi, mites and bacteria These organisms infect the grains and degrade the quantity and quality of the grains.
2. Abiotic factors: Various factors, like inappropriate moisture and temperature, can lead to the loss of grains and their discolouration, loss in weight, etc.

Question 6.
How do good animal husbandry practices benefit farmers?
Answer:
Good animal husbandry practices produce draught animals that can be used for the agricultural purposes like tilling, irrigation and carting. The draught animals perform these works and make the work easy for the farmer.

Question 7.
What are the benefits of cattle farming?
Answer:
Cattle farming is beneficial in the following ways:

1. Milk production is increased by high – yielding animals.
2. Good quality of meat, fibre and skin can be obtained.
3. Good breed of draught animals can be obtained.

Question 8.
For increasing production, what is common in poultry, fisheries, and bee-keeping?
Answer:
Proper management technique is a common factor in increasing production in poultry, fisheries, and beekeeping. Management techniques like regular cleaning of farms are of utmost importance. Maintenance of temperature, prevention, and cure of diseases of the farmed organisms are also required to increase the number of animals and the yield.

Question 9.
How do you differentiate among capture fishing, mariculture and aquaculture?
Answer:

1. Capture fishing is the method of obtaining fishes from natural resources.
2. Mariculture is the culture of marine fishes for commercial use. It is the farming of marine fishes of high economic value in seawater. For example, the farming of finned fishes like mullets, bhetki, pearl spots, etc.
3. Aquaculture involves the production of aquatic animals that are of high economic value such as prawns, lobsters, fishes, crabs, etc.

JAC Class 9 Science Solutions